Showing $|f^{(n)}(0)|\leq (n+1)!e.$

Question

Let $f(z)\in H(D)$ and suppose that $$|f(z)|\leq\frac{1}{1-|z|}\:\:\: \text{for all }\:\:z\in D.$$

Show that

$$|f^{(n)}(0)|\leq (n+1)!e.$$

Applying Cauchy integral formula I get the following inequality, however I don't know how to obtain $(n+1)!e$ as an upper bound. Thank you!

$$|f^{(n)}(0)|=\frac{n!}{2\pi }|\int_{D}\frac{f(z)}{z^{n+1}}dz|$$ $$\leq \frac{n!}{2\pi }\int_{D}|\frac{f(z)}{z^{n+1}}|.|dz|$$ $$\leq\frac{n!}{2\pi }\int_{D}\frac{1}{|z^{n+1}|(1-|z|)}.|dz|. $$

Answer 1

You can see with Cauchy integral formula, for $r<1$ $$f^{(n)}(0)=\dfrac{n!}{2\pi i}\int_{|z|=r}\dfrac{f(z)}{z^{n+1}}dz$$ and $z=re^{i\theta}$ $$|f^{(n)}(0)| \leqslant \dfrac{n!}{2\pi r^n}\int_0^{2\pi}|f(re^{i\theta})|d\theta \leqslant \dfrac{n!}{2\pi r^n}\int_0^{2\pi}\dfrac{1}{1-r}d\theta = \dfrac{n!}{r^n(1-r)}$$ The function $g(r)=\dfrac{1}{r^n(1-r)}$ is maximum in $r=\dfrac{n}{n+1}$ so with substitution you find $$|f^{(n)}(0)| \leqslant n!(n+1)\left(1+\dfrac1n\right)^n\leqslant(n+1)!e$$